# I want to know how to determine det(B) if det(A) = 3? 1 a 1 4 2a+1 2+b A      = 2 1 b B   = 15 5c 0...

I want to know how to determine det(B) if det(A) = 3?

 1 a 1 4 2a+1 2+b A      = 2 1 b B   = 15 5c 0 3 C 0 2 1 b

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

First we will write the formula for det (A)

det(A) = 1*1*0 + a*b*3 + 2*c* *1 - 1*1*3 - a*2*0 -c *b*1 = 3

Now simplify:

det(A) = 3ab + 2c - 3 -bc = 3

= 3ab + 2c - bc = 6............(1)

Now we will calculate det(B):

det(B) = 4*5c*b + (2a+1)*0*2 + (2+b)*15*1 - (2+b)*5c*2 - (2a+1)*15*b - 4*0*1

= 20bc + 30 + 15b - 20c - 10bc - 30ab - 15b

= 10bc - 20c + 30 -30ab

Let us rewrite:

==>det(B) = -30ab - 20c + 10bc + 30

Factor = -10:

==> det (B) = -10 (3ab + 2c - bc) + 30

But from (1) we know that (3ab+2c - bc) = 6

==> det(B) = -10*6 + 30

= -60 + 30 = -30

==> det(B) = -30

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Expandinfg thruogh the 1st row, we get :

Det (A) =  |[(1  a  1),(2  1  b), (3  c  0)]| = 3......(1)

Det B = |[(4  2a+1  2+b ) , (15  5c  0) , (2 1  b)]|.

Det B = R1-R3 = |[(4-2  2a+1-1  2+b-b) , (15  5c  0), (2  1  b)]|.

DetB = |[(2  2a  2), (15   5c  0) , (2  1   b)]|

Det B = 2 *5 |[(1  a  1), (2  c  0) , (2  1  b)]| , where 2 and 5 are factored out from 1st and 2nd rows.

Now we inter change R2 and R3  which results

Det B = -10 |[( 1  a  1), (2   1   b) , ( 2   c   0)]|.....(2)

Now we notice from the right side of  eq (1) and  left eq (2) that:

Det (B) = -10 det (A). But det(A) = 3.

Therefore det(B) = -10*3 = -30.

Therefore det(B) = -30